Back to: O level chemistry notes full Uganda syllabus

Mole concept deals with determining or counting the number of particles. Since the number of particles is in large quantities, it becomes very difficult to deal with large numbers; therefore, these particles are placed in groups called moles. One mole of a substance has 6.02×1023 particles. The particles of substances grouped into moles can be molecules, atoms, ions, electrons, radicals, protons or any other specified particles. For example 1 mole of magnesium atoms contains 6.02×1023 atoms, 1 mole of magnesium ions contains 6.02×1023 ions, 1 mole of H2SO4(aq) molecules contains 6.02×1023 molecules of H2SO4(aq).

A mole is the amount of substance which contains 6.02×1023 particles. Or

A mole is the amount of substance that contains the same number of particles as the number of particles in 12grams of carbon-12.

The number 6.02×1023 is called Avogadro‘s number or constant and it is denoted by letter L.

Molar Mass

Molar mass is the mass of one mole of a substance. It is equal to the relative atomic mass expressed in grams. The relative atomic masses of all elements have already been established. Examples are given below.

**Formula mass or molecular mass**

This is the mass of one mole of a compound. It is obtained by adding the relative atomic masses of the atoms present in a compound. The formula mass of a compound is equal to the relative formula mass expressed in grams.**Examples**

Calculate the formula masses of the following compounds.

**Exercise**

- Calculate the following (Al=27,H=1,O=16,S=32,Ca=40,Na=23,Hg=201,Cl=35.5)

a) Number of atoms in 2 moles of sodium

Steps in calculating empirical formulae

- Write down the symbols of the elements present
- Write down the percentage composition or composition by mass below the symbols
- Find the number of moles of each element by dividing the percentage composition or mass by Relative Atomic Mass
- Find the mole ratio of the elements by dividing the moles with the smallest number
- Write down the empirical number.

If the mole ratio is in fractions; - Round off to the nearest whole number if it is very close to the whole number.

- Multiply by a small number that converts the fraction to a whole number if the fraction is not close to a whole number.
**Molecular formula**

Is a formula that shows the actual number of each atoms present in one molecule of a compound. The molecular formula is a multiple of the empirical formula, so, from the empirical formula, the molecular formula can be determined.

Molecular formula= (Empirical formula)n = Molecular mass

n is number to be determined

Calculations on empirical and molecular formulae**Examples** - a) Calculate the empirical formula of a compound containing 80% carbon and 20% hydrogen.

b) If the molar mass of the compound is 30g, determine its molecular formula.

**Exercise**

- A compound X consists of carbon 40%, hydrogen 6.7% and the rest being oxygen. If the RMM is 60, determine its molecular formula.(C=12,H=1,0=16)(Ans. CH2O)
- A hydrocarbon is made up of 92.3% carbon and has molecular formula of 78g. Calculate its empirical and molecular formula.(Answer CH)
- Calculate the empirical formula of the compound formed when 1.8g of carbon forms 2.4g of a hydrocarbon. (Answer CH4)
- Calculate the molecular formula of a hydrocarbon with empirical formula CH2 and molecular mass of 28g. (Answer C2H4)
- Calculate the empirical formula of a salt with the following composition, copper 25%, sulphur 12.8%, oxygen 25.6% and water 36.0% (Answer CuSO4.5H2O)
- Calculate the empirical formula of a hydrated salt with the following composition, sodium 16.09%, carbon 9.20%, oxygen 16.78% and water 62.93% (Answer Na2C2O3.10H2O)
- Find the empirical formulae of the compounds formed in the reactions described below.

a) 10.800g magnesium form 18.000g of an oxide (Answer=MgO)

b) 3.400g calcium form 9.435g of a chloride (Answer=CaCl2)

c) 3.528g iron form 10.237g of a chloride.( Answer=FeCl3) - Calculate the empirical formulae of the compounds from which the following analytical results were obtained.

a) 27.3%C, 72.7%O (Answer=CO2)

b) 53.0%C, 47.0%O (Answer=C3O2)

c) 29.1%Na, 40.5%S,30.4%O (Answer=Na2S2O3)

d) 32.4%Na, 22.5%S, 45.0%O (Answer=Na2SO4)**Calculation of masses from equations**

Moles and mole ratios can be used to calculate the amount of substances reacting and products formed. This requires that a correctly balanced equation is written. Such an equation is known as a stoichiometric equation. Stoichiometry is the relationship between amounts of reactants and products in a chemical reaction.

A stoichiometric equation is an equation in which the reactants and products are correctly balanced.

Steps involved in the calculation - Write down a balanced equation for the reaction

- Convert the moles into grammes

Examples - Calculate the mass of iron (II) sulphide formed by heating 64g of sulphur with excess iron filling.(S=32,Fe=56)
- Write down the moles of substances that concerns the question

**Exercise**

- A solution of 8.1g of NaOH was neutralized by hydrochloric acid. Calculate the mass of sodium chloride produced when the solution was evaporated to dryness.

(C=12, Na=23, O=16, H=1, Cl=35.5,) (Answer =11.85g) - Calculate the mass of residue left when 2.40g of sodium hydrogen carbonate is decomposed by heat. (Answer =1.51g)
- Calculate the loss in mass when 100g of calcium carbonate is heated to constant mass. (Ca=40, C=12, 0=16) (Answer =44g)
- 76.5g of calcium hydrogen carbonate was heated strongly. What was the mass of carbon dioxide formed? (Answer =20.78g)
- What mass of sodium oxide would be made from 1.5 g of sodium? (Answer=2.02g)